At S.T.P. the volume of $7.5 \mathrm{g}$ of a gas is $5.6 \mathrm{L}$. The gas is

Given, mass of gas $(W)=7.5 \mathrm{g}$ Volume of gas at $\operatorname{STP}(V)=56 \mathrm{L}$ $\because$ Moles of gas at $\mathrm{STP}=\frac{V(\mathrm{L})}{22.4(\mathrm{L})}=\frac{W}{M}$ Where $M=$ molar mass of gas. $\therefore \quad M=\frac{W \times 22.4}{V}$ $$ =\frac{7.5 \times 22.4}{5.6}=30.00 \mathrm{g} \mathrm{mol}^{-1} $$ Among the given options Molar mass $(M)$ of $\mathrm{NO}=14+16=30.00 \mathrm{g} \mathrm{mol}^{-1}$ (b) $\mathrm{N}_{2} \mathrm{O}=28+16=44.00 \mathrm{g} \mathrm{mol}^{-1}$ (c) $\mathrm{CO}=12+16=28.00 \mathrm{g} \mathrm{mol}^{-1}$ (d) $\mathrm{CO}_{2}=12+32=44.00 \mathrm{g} \mathrm{mol}^{-1}$ $\because$ Molar mass of $\mathrm{NO}=30 \mathrm{g} \mathrm{mol}^{-1}$ Thus, the given gas is NO. Hence, is the correct option.